# The volume of a cone is $18π m^3$ Find the minimum length of the slant edge

Using pythagoras theorem, I received.. $$l (Slant)=\sqrt{r^2+h^2}$$ Using the volume of a cone formula in terms of $h$..$$h=\dfrac{54}{r^2}$$ I then subbed this into the 1st equation and diffrenciated with respect to $r$..$$\dfrac{\mathrm dl}{\mathrm dr}=\dfrac{2r-\dfrac{11664}{r^5}}{2\sqrt{(r^2+\dfrac{2916}{r^4})}}$$

Looking at it now I probably should have replaced 54 with a symbol, can someone tell me if I'm on the right track? I put it equal to zero then transposed in terms of $r$ and received $1.45$ for the radius. I subbed that value back into the $l$ equation on top and recieved the wrong answer...the answer is meant to be $3\sqrt{3}$

$2r-\dfrac{4*54^2}{r^5}=0$ means $r^6=2*54^2=2^3*(3^2)^3$, $r^2=2*3^2=18$

Hence $h^2=9$ and $l^2=18+9$, $l=3\sqrt 3$.

• I made a silly error, typed into my calculator incorrectly..thanks for the answer! Jun 17 '15 at 11:24
• @Modrisco You should avoid if possible to use calculator! Usually problems are made to get "nice" results, and this means simplifications to look for around. Jun 17 '15 at 11:55

$$l^2=r^2+\left(\dfrac{54}{r^2}\right)^2=\dfrac{r^2}2+\dfrac{r^2}2+\left(\dfrac{54}{r^2}\right)^2$$

Now using AM-GM inequality for three variables,

$$\dfrac{\dfrac{r^2}2+\dfrac{r^2}2+\left(\dfrac{54}{r^2}\right)^2}3\ge\sqrt{\dfrac{r^2}2\cdot\dfrac{r^2}2\cdot\left(\dfrac{54}{r^2}\right)^2}$$

The equality occurs if $\dfrac{r^2}2=\dfrac{r^2}2=\left(\dfrac{54}{r^2}\right)^2$

As $r\ge0,$ we need $\dfrac r{\sqrt2}=\dfrac{54}{r^2}\iff r=3\sqrt2$

• $$\implies\dfrac{l^2}3\ge \sqrt{27^2}=(3^6)^{1/3}=9$$ Jun 17 '15 at 11:08
• AM-GM inequality for 3 variables? I'm not sure what's going on here...is there another more simpler way? I'm not that advanced in mathematics. Jun 17 '15 at 11:18
• @Modrisco, See en.wikipedia.org/wiki/… This helps you avoiding calculus Jun 17 '15 at 11:26